Square root of radicals Find the square root of $4^{1/3}+16^{1/3}+1$.
I tried to solve by supposing the square root to be $x$ and then cubing both sides but it didn't work. 
I do not need exact value. 
By hit and trial
I have seen that answer should involve $2^{1/3}+1....$ 
Thanks 
 A: I will assume you are interested only in real numbers and not complex numbers.  If you wish to simplify so that there are no nested radicals:
First express inner radicals as exponentials terms of the same base
$\sqrt{1+\sqrt[3]{4}+\sqrt[3]{16}}=\sqrt{1+2^{2/3}+2^{4/3}}=\sqrt{1+2\cdot 2^{1/3}+2^{2/3}}$
Above, we used that $2^{4/3}=2^{3/3}\cdot 2^{1/3}$.
Now, let $x=2^{1/3}$.  We recognize that the above can be written as
$\sqrt{1+2x+x^2}=\sqrt{(1+x)^2}=|1+x|$.  Remembering that $\sqrt[3]{2}$ is a positive real, we may replace this back in for $x$ and remove the absolute value sign to get that the original expression is equal to $1+\sqrt[3]{2}$
A: Notice that $4^{1/3}=2^{2/3}=\left(2^{1/3}\right)^2$ and $16^{1/3}=(8\cdot2)^{1/3}=2\cdot2^{1/3}$. Therefore:
$$4^{1/3}+16^{1/3}+1=\left(2^{1/3}\right)^2+2\cdot2^{1/3}\cdot1+1^2=\left(2^{1/3}+1\right)^2.$$
A: $$16^{1/3}=2^{1/3} . 8^{1/3}=2 . 2^{1/3}$$
So we have $$2^{{1/3}^2} +2 . 2^{1/3}+1 = (2^{1/3}+1)^2$$
So the root is $$2^{1/3} +1$$
A: $1^{2/3},2^{2/3},4^{2/3}$ is GP with $a=1^{2/3}, r=2^{2/3}, n=3$
$\therefore S=1^{2/3}\cdot \frac{(2^{2/3})^3-1}{2^{2/3}-1}=\frac{3}{2^{2/3}-1}$
$\therefore$ the required answer$=\sqrt{\frac{3}{2^{2/3}-1}}$
